Question

When a cricket ball is thrown vertically upwards with an acceleration of 10m/s^2 reaches a maximum height of 5 m ?what was the initial speed of the ball ? How much time is taken by the ball to reach the highest point ?

Answer 1

Answer:

Explanation:

Take equation'. 2as=v^2-u^2'

V=final velocity=0(because ball will stop at highest point)

U=initial velocity

2×10×5=0^2-u2

100=0-u^2

U^2=100

U=10m/s

Initial speed of ball =10m.s

Answer 2

Answer:

Explanation:

S= 5m

v=0

u=u

a=10m/s

putting the values in equation

v^= u^2 +2as

0 = u^2 + 2 X 10 X 5

u = -10m( where negative sign indicates direction)

(2) now putting the values in equation

v = u + at

0 = -10 + 10 X t

t = 1 sec